EMG 211: ENGINEERING MATHEMATICS I
COURSE OUTLINES PART ONE
• • • • Maxima and Minima of Functions of a Single Independent Variable Tangents and Normals Differentiation Techniques of Differentiation
• Techniques of Integration: Indefinite Integrals, Integration by Parts, Definite Integrals, Improper Integrals • • Applications to Engineering Systems Introduction to Ordinary Differential Equations (ODE) and Partial Differential Equations (PDE)
• • • Properties and Evaluation of Matrices Introduction to Symmetric and Skew-symmetric Matrices Simple simultaneous Linear Equations
Maxima and Minima, Tangents and Normals
Gradients: The gradient of a straight line is the rate of change of y with x and is measured by =
the increase in the value of y divided by the corresponding increase in the value of x. Thus, if
= the tangent at that point and when the equation of the curve is known differentiation provides a method of calculating the gradient exactly. Example: Calculate the gradient of the curve Solution:
1.2 =9 − 4 + 5. When x = 3, =3 −2 + 5 − 7 at the point (3, 71). + , = gradient of the line. The gradient of a curve at a particular point is the gradient of
= 74 = gradient of the curve.
Tangents and Normals: For the curve given by
gives the gradient of the at that point,
tangent at any point. Given the co-ordinates of a point on the curve and the value of the equation of the tangent can be easily obtained. Example: Find the equation of the tangent to the curve Solution:
= + 3 + 2;
=3 = + 3 + 2 when x = – 2
Therefore the required equation is
= 15 + 18
+ 3. When x = – 2,
= 15. Also, when x = – 2, y = – 12.
− 15 − 18 = 0.
A normal is a line perpendicular to the tangent. Recall from geometry that if two lines are perpendicular then the product of their gradients is – 1. Example: Calculate the equation of the normal for the previous example at the points already defined. and equation of the normal is 15 + is parallel to the x-axis. Solution: The point is (– 2, – 12); the gradient of the tangent is 15. :. Gradient of the normal =
− + 182 = 0. = −4 ,
Example: Find the coordinates of the points on the curve
− 3 − 2 where the tangent
Solution: A line parallel to the x-axis has zero gradient. If the tangent is parallel to the axis of x then = 0 at the point of contact. =3 − 8 − 3 = (3 + 1)( − 3) = 0. :. = −1⁄3
are the only values of x which give points on the curve where the gradient is 0. Their coordinates are
(− 1⁄3 , − 40⁄27) and (3, – 20). 1.3 Stationary Points
A point on a curve at which
= 0 is called a stationary point and the value of the function at that
point is called a stationary value. At such points the tangent is parallel to the x-axis. To find the stationary points, set = 0 and face the consequence. + 15 + 18 + 7.
Example: Find the stationary points of the function
4 Solution: When = 12 + 30 + 18 = 6(2
+ 5 + 3) = 6(x + 1) (2x + 3)
= 0; 6(x + 1) (2x + 3) = 0, giving
= – l or – 3/2. The stationary points are therefore (– l, 0)
and (– 3/2, 1/4) and the stationary values of the function at these points are 0 and 1/4, respectively. 1.4 Maxima and Minima
Figure 1.1 shows a curve passing through a stationary point and reaching a maximum value at that point. As increases (from left to right) the gradient of the curve decreases from a positive value
through 0 to a negative value. Figure 1.2 shows a curve reaching a minimum at a point. As increases the gradient increases from a negative value through 0 to a positive value. Maximum and minimum points are also referred to as Turning Points (TP) because the tangent turns over at such points. For Figures 1.1 and 1.2, please come to class. 1.5 Points of Infexion
A third type of stationary point is shown in Figure 1.3, where the curve has neither a maximum nor a minimum value. This is called a point...
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